Question

In a study of proteins mediating cell membrane transport, microbiologists measure current versus time through the cell membranes of oocytes (nearly mature egg cells) taken from the African clawed frog, Xenopus. The measured current versus time is given approximately by I=60t+200t^2+4.0t^3, with t in seconds and I in nA. Find the total charge that flows through the cell membrane in the interval from t=0 to t=5.0s

Answer 1

Answer:

The total charge that flows in the 5 second interval is 12,800 nA

Explanation:

To calculate the total charge, we need to find out the charge at each one second intervals, In order to do that, just plug in the value of t into the expression to find the value

I = 60(t) + 200(t²) + 4(t³)

at t = 0

I = 60(0) + 200(0²) + 4(0³) = 0 nA

at t = 1

I = 60(1) + 200(1²) + 4(1³) = 264 nA

at t = 2

I = 60(2) + 200(2²) + 4(2³) = 952 nA

at t = 3

I = 60(3) + 200(3²) + 4(3³) = 2088 nA

at t = 4

I = 60(4) + 200(4²) + 4(4³) = 3696 nA

at t = 5

I = 60(5) + 200(5²) + 4(5³) = 5800 nA

Now add all of them together

0 + 264 + 952 + 2088 + 3696 + 5800 = 12,800 nA

Hope that answers the question. Have a great day!

Answer 2

Answer:

Q(0) = 0C, Q(1) = 264nC, Q(2) = 952C Q(3) = 2088nC, Q(4) = 3696C Q(5) = 5800nC

Explanation:

I = 4t³ + 200t² + 60t

But charge of an object =》 Q = IT

Charge of an object is the product of the current and the time in which the current passes through the membrane.

When t = 0

Q = 4(0)³ + 200(0)² + 60(0) = 0C

When t = 1

Q = 4(1)³ + 200(1)² + 60(1) = 264nC

When t = 2

Q = 4(2)³ + 200(2)² + 60(2) = 952nC

When t= 3

Q = 4(3)³ + 200(3)² + 60(3) = 2088nC

When t= 4

Q = 4(4)³ + 200(4)² + 60(4) = 3696nC

When t = 5

Q = 4(5)³ + 200(5)² + 60(5) = 5800nC